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Chapter 6: Problem 15

It takes \(185 \mathrm{kJ}\) of work to accelerate a car from \(23.0 \mathrm{m} / \mathrm{s}\) to \(28.0 \mathrm{m} / \mathrm{s} .\) What is the car's mass?

Short Answer

Expert verified
The car's mass is approximately 1450 kg.

Step by step solution

01

Understanding the Problem

In this problem, we need to find the car's mass based on the work done to change its velocity. We know the initial and final velocities, as well as the work done on the car. We'll use the work-energy principle.
02

Applying Work-Energy Principle

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. So, we can write: \[ W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \] where \( W = 185 \text{kJ} = 185,000 \text{J} \), \( u = 23.0 \text{ m/s} \), and \( v = 28.0 \text{ m/s} \).
03

Setting Up the Equation

Using the formula from the previous step, set up the equation substituting the known values: \[ 185,000 = \frac{1}{2} m (28.0)^2 - \frac{1}{2} m (23.0)^2 \] Simplify the equation: \[ 185,000 = \frac{1}{2} m (784 - 529) \] which gives us \[ 185,000 = \frac{1}{2} m \times 255 \]
04

Solving for Mass

To isolate the mass \( m \), multiply both sides by 2 to get rid of the fraction: \[ 370,000 = 255m \] Now, solve for \( m \) by dividing both sides by 255: \[ m = \frac{370,000}{255} \] which results in \( m \approx 1450.98 \text{ kg} \).
05

Final Calculation Check

Re-calculate to ensure no errors: \[ m = \frac{370,000}{255} = 1450.98 \text{ kg} \] Round to the correct number of significant figures based on the given data, which are usually three significant figures from measurements: \( m \approx 1450 \text{ kg} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. This type of energy depends on two key factors: the mass of the object and its velocity. The formula for calculating kinetic energy (KE) is:\[ KE = \frac{1}{2}mv^2 \]where:
  • \( m \) is the object's mass in kilograms (kg)
  • \( v \) is the velocity in meters per second (m/s)
For example, if a car speeds up from one velocity to another, its kinetic energy changes according to the velocities. This change is fundamental when applying the work-energy principle to find other variables like mass, as shown in our exercise.
Work Done
Work done in physics refers to the transfer of energy that occurs when a force makes an object move. In the context of kinetic energy, the work done is the change in kinetic energy. The equation is:\[ W = \Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} mu^2 \]where:
  • \( W \) is the work done in joules (J)
  • \( \Delta KE \) is the change in kinetic energy
  • \( u \) and \( v \) are initial and final velocities, respectively
In our exercise, the work done was provided as \(185 \text{kJ}\), which is \(185,000 \text{J}\). This work is responsible for accelerating the car and changing its kinetic energy.
Velocity Change
Velocity change is a vital concept in understanding the shift in an object's motion, which directly affects kinetic energy and work calculations. Velocity (\( v \)) refers to the speed of an object in a particular direction.
In our problem, the car goes from an initial velocity of \(23 \text{ m/s}\) to a final velocity of \(28 \text{ m/s}\). The difference in velocities impacts the kinetic energy change by affecting the calculation \(v^2 - u^2\). The greater the change in velocity, the larger the shift in kinetic energy, which corresponds to the work done on the object.
Mass Calculation
Mass calculation using the work-energy principle involves isolating the variable \( m \) from the equations for work done and kinetic energy change.
From the formula:\[ W = \frac{1}{2} m (v^2 - u^2) \]where \( W = 185,000 \text{ J} \), \( v^2 = 784 \), and \( u^2 = 529 \), we simplify it to:\[ 185,000 = \frac{1}{2} m (784 - 529) \]Solving for \( m \):
  1. Calculate \( v^2 - u^2 = 255 \)
  2. Multiply both sides by 2: \( 370,000 = 255m \)
  3. Divide both sides by 255 to find \( m \)
This yields an approximately calculated mass of \(1450 \text{ kg}\). This method provides a clear way to derive mass from changes in kinetic energy due to work done.

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Most popular questions from this chapter

6\. A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of \(22.0 \mathrm{m} .\) She pushes in a direction \(29.0^{\circ}\) below the horizontal. A \(48.0-N\) frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

A 75.0-kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of \(14.6^{\circ}\) with the horizontal. What is the change in the skier's gravitational potential energy?

A bicyclist rides 5.0 km due east, while the resistive force from the air has a magnitude of \(3.0 \mathrm{N}\) and points due west. The rider then turns around and rides \(5.0 \mathrm{km}\) due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of \(3.0 \mathrm{N}\) and points due east. (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

Two pole-vaulters just clear the bar at the same height. The first lands at a speed of \(8.90 \mathrm{m} / \mathrm{s}\), and the second lands at a speed of \(9.00 \mathrm{m} / \mathrm{s}\). The first vaulter clears the bar at a speed of \(1.00 \mathrm{m} / \mathrm{s}\). Ignore air resistance and friction and determine the speed at which the second vaulter clears the bar.

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A \(70.3-\mathrm{kg}\) water-skier has an initial speed of \(6.10 \mathrm{m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{m} / \mathrm{s} .\) Determine the work done by the net external force acting on the skier.
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